Finding analytic function from imaginary part using Cauchy-Riemann equations

[dan280291]

Homework Statement

Given v(x,y) find $f(z) = u(x,y) +iv(x,y)$
v(x,y) = 3y -2(x^2 - y^2) +(x) / (x^2 + y^2)

The Attempt at a Solution

Using Cauchy Riemann relations I've found

$dv/dx = -4x + (x^2+y^2)^-1) +2x^2(x^2+y^2)^-2 = -du/dx$

Now integrate that with respect to y to find u

But I'm not too sure how to integrate the fractions partially.

Also I've found $dv/dy = 3 +4y -2yx/(x^2 + y^2)$

 

[mfb]

But I'm not too sure how to integrate the fractions partially.

Where is the problem? You have to find a function u, that, when you calculate the derivative with respect to x, has to have some specific form, and for the derivative with respect to y you get another known expression (check your relation(s)).
To find u, you can use an integration with respect to x or y, respectively.

 

[dan280291]

Hi thanks for the reply. Does that mean you only have to find dv/dy and integrate with respect to x? You don't need to use dv/dx?

Also as for the integration how is 2yx/(x^2 + y^2) integrated with respect to x?

Thanks again.

 

[Ray Vickson]

Hi thanks for the reply. Does that mean you only have to find dv/dy and integrate with respect to x? You don't need to use dv/dx?

Also as for the integration how is 2yx/(x^2 + y^2) integrated with respect to x?

Thanks again.

You need both $\partial u/\partial x = \partial v/ \partial y$ and $\partial u/ \partial y = - \partial v / \partial x$.

 

[dan280291]

Thanks for the help off everyone. Last thing could anyone tell me how 2yx/(x^2 + y^2) is integrated with respect to x?

 

[mfb]

Hint: look at the derivative of the denominator. Do you see some nice substitution?